At \(s=1\) the generating function is the total mass of \(\theta \), while at \(s=0\) only the constant term remains. If \(0\leq s\leq t\), then \(s^j\leq t^j\) for every \(j\), and multiplication by the nonnegative weight \(\theta _j\) followed by summation proves monotonicity. Taking \(t=1\) shows that \(f(s)\leq 1\) on \([0,1]\), and nonnegativity follows in the same way from the nonnegativity of every summand. Since the sum is finite, we may differentiate it term by term. Evaluating the resulting derivative at \(1\) gives \(f'(1)=\sum _j j\theta _j=m\).

Let \(A=\{ s\in [0,1]\colon f(s)=s\} \). The set \(A\) is nonempty because \(1\in A\), bounded below by \(0\), and closed because \(f\) is continuous. A nonempty closed subset of \(\mathbb {R}\) that is bounded below contains its infimum, so \(q\in A\). This gives both \(q\in [0,1]\) and \(f(q)=q\). If \(s\in [0,1]\) is any other fixed point, then \(s\in A\), and the defining property of the infimum gives \(q\leq s\).

If \(q=0\), then \(\theta _0=f(0)=f(q)=q=0\). Conversely, if \(\theta _0=0\), then \(f(0)=0\), so \(0\) is a fixed point in \([0,1]\). The minimality in lemma 2.5 gives \(q\leq 0\), while \(q\geq 0\), and hence \(q=0\).

Write

The polynomial \(g\) satisfies \(g(1)=m{\gt}1\). By continuity, we may choose \(s\in (0,1)\) such that \(g(s){\gt}1\). Since \(1-s{\gt}0\), the displayed identity gives \(f(s){\lt}s\). The continuous function \(f(t)-t\) is nonnegative at \(0\) and negative at \(s\), so it has a zero \(c\in [0,s]\). By lemma 2.5, \(q\leq c\leq s{\lt}1\).